hw27solutions - the endpoints. Since f is bounded, there...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: April 21, 2010 Assignment 27 1. If f is continuous on [ a,b ], a < b , show that there exists c [ a,b ] such that we have R b a f ( x ) dx = f ( c )( b - a ). This result is sometimes called the Mean Value Theorem for Integrals . This is most easily done using the Fundamental Theorem of Calculus and the Mean Value Theorem. Indeed, set F ( x ) = R x a f ( y ) dy . Then, the Fundamental Theorem of Calculus says that F is differentiable. Thus, there is a c [ a,b ] so that F ( b ) - F ( a ) = F 0 ( c )( b - a ) . But this is equivalent to Z b a f ( x ) dx = f ( c )( b - a ) . 2. If f is bounded and there is a finite set E such that f is continuous at every point of [ a,b ] \ E , show that f ∈ R [ a,b ]. It suffices to prove this when E contains only the point b . The proof for when E = { a } is quite similar. And for any other set of E , we can use additivity to break the desired result into proving integrability with the only discontinuity being at one of
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Unformatted text preview: the endpoints. Since f is bounded, there exists m,M R so that m f ( x ) M for all x [ a,b ]. Let &gt; 0 be given. We set = / 2( M-m ). Since f is continuous on [ a,b- ], there is a partition P of [ a,b- ] so that U ( P,f )-L ( P,f ) &lt; / 2 . As a partition for [ a,b ], we set P = P { b } . Then, U ( P,f )-L ( P,f ) = U ( P,f )-L ( P,f ) + ( M *-m * ) where M * ,m * are the sup and inf of f over [ b-,b ] respectively. Then, U ( P,f )-L ( P,f ) &lt; ( / 2) + ( M-m ) = . Since &gt; 0 was arbitrary, this shows that f R [ a,b ] as desired. 1...
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